Google
Câu hỏi: Cho $\int{f\left( x \right)\text{d}x={{x}^{2}}-3x+C}$. Tìm $\int{f\left( {{e}^{-x}} \right)\text{d}x}$.
A. $\int{f\left( {{e}^{-x}} \right)\text{d}x}={{e}^{-2x}}-3{{e}^{-x}}+C$.
B. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=2{{e}^{-x}}-3x+C$.
C. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=-2{{e}^{-x}}-3x+C$.
D. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=-2{{e}^{-x}}-3{{e}^{-x}}+C$.
A. $\int{f\left( {{e}^{-x}} \right)\text{d}x}={{e}^{-2x}}-3{{e}^{-x}}+C$.
B. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=2{{e}^{-x}}-3x+C$.
C. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=-2{{e}^{-x}}-3x+C$.
D. $\int{f\left( {{e}^{-x}} \right)\text{d}x}=-2{{e}^{-x}}-3{{e}^{-x}}+C$.
Từ giả thiết $\int{f\left( x \right)\text{d}x={{x}^{2}}-3x+C}\Rightarrow f\left( x \right)=2x-3\Rightarrow f\left( {{e}^{-x}} \right)=2{{e}^{-x}}-3$
Khi đó $I=\int{f\left( {{e}^{-x}} \right)\text{d}x}=\int{\left( 2{{e}^{-x}}-3 \right)\text{d}x}=-2{{e}^{-x}}-3x+C$.
Khi đó $I=\int{f\left( {{e}^{-x}} \right)\text{d}x}=\int{\left( 2{{e}^{-x}}-3 \right)\text{d}x}=-2{{e}^{-x}}-3x+C$.
Đáp án C.